Answer
$\sin\theta$ = $\frac{\sqrt 3}{2}$
Work Step by Step
We know from first Pythagorean identity that-
$\sin\theta$ = ± $\sqrt (1-\cos^{2}\theta)$
As $\theta$ terminates in Q II, Therefore $\sin\theta$ will be positive-
$\sin\theta$ = $\sqrt (1-\cos^{2}\theta)$
substitute the given value of $\cos\theta$-
$\sin\theta$ = $\sqrt (1-(\frac{-1}{2})^{2})$
$\sin\theta$ = $\sqrt (1-\frac{1}{4})$
$\sin\theta$ = $\sqrt (\frac{4-1}{4})$ = $\sqrt (\frac{3}{4})$
$\sin\theta$ = $\frac{\sqrt 3}{2}$
